Show how to use an infinite series of matrix powers to compute $A^{-1}$ for $A \in \mathbb{R}^{n\times n}$. Which conditions must be satisfied by the eigenvalues of $A$ for the series to converge?
Asked
Active
Viewed 464 times
1
-
A hint to get you started: for $x\in (-1,1)$ we have $\frac{1}{1-x} = \sum_{n=0}^\infty x^n$. – James Nov 07 '17 at 15:50
-
1I already found this explatation. So I need to find a $B$ so that $A=(I-B)$, $|B|<1$ and then the cauchy sequence converges to $A^{-1}$. I am just not sure how to formaly proof this. – MathSchoolar Nov 07 '17 at 16:04
-
@Math1000 I think you meant the reverse inequality. – Federico Poloni Nov 07 '17 at 16:32
-
Indeed, $|A|\geqslant |\lambda|$ for any eigenvalue $\lambda$. – Math1000 Nov 07 '17 at 16:52
-
@Math1000 where does this come from? This seems to contradict the information here – MathSchoolar Nov 07 '17 at 17:47
-
If $|x|=1$ and $Ax=\lambda x$, then $|Ax| = |\lambda x| = |\lambda|$... – Math1000 Nov 07 '17 at 18:22